The time required to cook a pizza at a neighborhood pizza joint is normally distributed with a mean of 12 minutes and a standard deviation of 2 minutes. Find the time for each event.

We are given that the time required to cook a pizza at a neighborhood pizza joint is normally distributed with a mean of 12 minutes and a standard deviation of 2 minutes.

Let X = the time required to cook a pizza at a neighborhood pizza joint.

The z-score probability distribution for the normal distribution is given by;

Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

where, [tex]\mu[/tex] = mean time = 12 minutes

[tex]\sigma[/tex] = standard deviation = 2 minutes

(a) We have to find the time for the event of the highest 5%, that means;

P(X > x) = 0.05 {where x is the required time}

P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{x-12}{2}[/tex] ) = 0.05

P(Z > [tex]\frac{x-12}{2}[/tex] ) = 0.05

Now, in the z table, the critical value of x that represents the top 5% of the area is given as 1.645, i.e;

[tex]\frac{x-12}{2}=1.645[/tex]

[tex]{x-12}=1.645\times 2[/tex]

x = 12 + 3.29 = 15.29 minutes.

Hence, the time for the event of the highest 5% is 15.29 minutes.

(b) We have to find the time for the event of the lowest 50%, that means;

P(X < x) = 0.50 {where x is the required time}

P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{x-12}{2}[/tex] ) = 0.50

P(Z < [tex]\frac{x-12}{2}[/tex] ) = 0.50

Now, in the z table, the critical value of x that represents the lowest 50% of the area is given as 0, i.e;

[tex]\frac{x-12}{2}=0[/tex]

[tex]{x-12}=0[/tex]

x = 12 + 0 = 12 minutes.

Hence, the time for the event of the lowest 50% is 12 minutes.

(c) We have to find the time for the event of the middle 95%, that means we have to find the time for the event of 2.5% and above 97.5%;

Firstly, the time for the event of the lowest 2.5%, i.e;

P(X < x) = 0.025 {where x is the required time}

P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{x-12}{2}[/tex] ) = 0.025

P(Z < [tex]\frac{x-12}{2}[/tex] ) = 0.025

Now, in the z table, the critical value of x that represents the lowest 2.5% of the area is given as -1.96, i.e;

[tex]\frac{x-12}{2}=-1.96[/tex]

[tex]{x-12}=-1.96 \times 2[/tex]

x = 12 - 3.92 = 8.08 minutes.

Firstly, the time for the event of the below 97.5%, i.e;

P(X < x) = 0.975 {where x is the required time}

P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{x-12}{2}[/tex] ) = 0.975

P(Z < [tex]\frac{x-12}{2}[/tex] ) = 0.975

Now, in the z table, the critical value of x that represents the top 2.5% of the area is given as 1.96, i.e;

[tex]\frac{x-12}{2}=1.96[/tex]

[tex]{x-12}=1.96 \times 2[/tex]

x = 12 + 3.92 = 15.92 minutes.

Hence, the time for the event of the middle 95% is 8.08 minutes and 15.92 minutes.

(d) We have to find the time for the event of the lowest 80%, that means;

P(X < x) = 0.80 {where x is the required time}

P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{x-12}{2}[/tex] ) = 0.80

P(Z < [tex]\frac{x-12}{2}[/tex] ) = 0.80

Now, in the z table, the critical value of x that represents the lowest 80% of the area is given as 0.8416, i.e;

[tex]\frac{x-12}{2}=0.8416[/tex]

[tex]{x-12}=0.8416 \times 2[/tex]

x = 12 + 1.68 = 13.68 minutes.

Hence, the time for the event of the lowest 80% is 13.68 minutes.